Geometric Distribution

A geometric setting requires: (B) Binary outcomes — each trial results in either 'success' or 'failure'. (I) Independent trials — the outcome of one trial does not affect any other. (N) No fixed number of trials — you keep going until the first success. (S) Same probability — the probability of success p is identical for every trial. When all four conditions hold, the random variable X = number of trials until the first success follows a geometric distribution.

To get the first success on exactly trial k, you need exactly (k − 1) failures followed by one success. Since trials are independent, you can multiply their probabilities. Each failure has probability (1 − p) and the success has probability p. Therefore P(X = k) = (1−p)^(k−1) · p, for k = 1, 2, 3, … This is a valid probability distribution because the sum over all k equals 1: the series ∑_{k=1}^{∞} p(1−p)^(k−1) = p · 1/(1−(1−p)) = 1.

The expected value (mean) of X is E[X] = 1/p. Intuitively, if each trial has a 20% chance of success, you expect to need 1/0.20 = 5 trials on average. The variance is Var(X) = (1−p)/p², and the standard deviation is σ = √((1−p)/p²). These formulas are derived using the geometric series and its derivatives. A smaller p means more trials are needed on average, and also greater spread in the distribution.

The geometric distribution is the only discrete distribution with the memoryless property: P(X > m + n | X > m) = P(X > n) for all positive integers m and n. In plain language, given that you have already had m failures, the probability of needing at least n more trials is exactly the same as if you were starting fresh. You can verify this algebraically: P(X > k) = (1−p)^k, so P(X > m+n | X > m) = P(X > m+n)/P(X > m) = (1−p)^(m+n)/(1−p)^m = (1−p)^n = P(X > n). ✓

Some textbooks define the geometric distribution as Y = number of failures before the first success (so Y = X − 1). In this version, Y can take values 0, 1, 2, … and P(Y = k) = (1−p)^k · p. The mean becomes E[Y] = (1−p)/p and the variance remains (1−p)/p². Both forms are valid; always check which definition a problem or formula sheet is using.

• X — Random variable (trials): The number of the trial on which the first success occurs; takes positive integer values 1, 2, 3, …
• k — Trial number (dimensionless): A specific positive integer representing which trial we are evaluating (k = 1, 2, 3, …)
• p — Probability of success (dimensionless): The constant probability of success on any single trial; must satisfy 0 < p ≤ 1
• (1-p)^{k-1} — Failure run probability (dimensionless): The probability of getting exactly k−1 failures in a row before the first success

Problem. A basketball player makes free throws independently with probability 0.70 each attempt. (a) What is the probability she first makes a free throw on her 3rd attempt? (b) What is the expected number of attempts until she makes her first free throw? (c) What is the probability she needs more than 4 attempts?

Answer. P(X = 3) = 0.063; E[X] ≈ 1.429 attempts; P(X > 4) = 0.0081

• Binomial Distribution — The binomial distribution counts successes in a fixed number of trials, while the geometric distribution counts trials until the first success — both arise from the same Bernoulli trial framework.
• Negative Binomial Distribution — The negative binomial generalises the geometric distribution by counting the number of trials needed to achieve r successes, making geometric distribution the special case r = 1.
• Geometric Series — The formula for the sum of a geometric series is used to prove that the geometric PMF sums to 1 and to derive the mean and variance formulas.
• Expected Value — The mean E[X] = 1/p of the geometric distribution is a key application of expected value for discrete random variables with infinite support.